does ringing at switch node in buck influences efficiency?

[batistuta]

hi, all
i want to know, in power saving mode ,does the ringing at switch node influences the buck's efficiency, i just know it will causes EMI.
can someone help me?
thanks

 

[VVV]

Re: does ringing at switch node in buck influences efficien

No, it does not influence the efficiency. The ringing is caused by the inductor resonating with the parasitic capacitance of the diode (and transistor). But that energy will be lost anyway, because the capacitance of the diode HAS to charge up to the output voltage. So if a capacitance charges up, it takes energy, E=0.5C*Vo^2. Anyway you look at it, it's lost energy.

 

[batistuta]

hi, vvv

does this energy will be used to charge the output capacitor? or lost by itself??

and if i use a switch connect to the two nodes of inductor, it works when the power mos all open, the switch will short the inductor, so the ringing will dispear, to decrease EMI issue.
does it can work well??

 

[VVV]

Re: does ringing at switch node in buck influences efficien

This energy is lost FROM the output capacitor and used to charge up the diode capacitance. Part of it will dissipate in the inductor. This all happens because when the diode turns off the voltage across it is zero, but its cathode (and hence one end of its capacitance) is connected to the output cap through the inductor.

So now you have the output cap charged to the output voltage and the diode capacitance, discharged, connected by an inductor. What's going to happen? The diode capacitance will charge up from the output capacitor through the inductor. Because the inductor has a low DC resistance (to keep losses low) the LC circuit that forms has a high Q, so it will "ring".

But no matter what, the output capacitor will lose a tiny bit of energy to charge up the diode capacitance. In the process this energy gets transferred back and forth between the diode capacitance and the inductor, creating ringing, but in the end it's still lost energy. And it does not matter where it gets dissipated.
Recall the physics problem with connecting two caps, one discharged and one charged to a certain voltage? In the end, there seemed to be some "missing" energy. In reality that energy did not dissapear, but was dissipated in the wires connecting the two caps. That energy was always the same, regardless of the resistance of the wire. That energy only depended on the capacitances in question and the voltages.
This is exactly the same situation here, except the "wire" is now the inductor, which, aside from dissiating some energy, also creates ringing in the process.

 

[nickwang]

Does anyone know how the ring would look like? For example, in a perfect L,Switch,Diode,Cap circuit, will the ringing always oscillating? Or will it damp? I'm seeing some strange waveform like a diode-modulated multi-frequency pulse. And it seems it will last forever.